Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $x \neq 0$. $z = \dfrac{-7x - 42}{x + 9} \times \dfrac{x^2 + 8x - 9}{3x + 18} $
Explanation: First factor the quadratic. $z = \dfrac{-7x - 42}{x + 9} \times \dfrac{(x + 9)(x - 1)}{3x + 18} $ Then factor out any other terms. $z = \dfrac{-7(x + 6)}{x + 9} \times \dfrac{(x + 9)(x - 1)}{3(x + 6)} $ Then multiply the two numerators and multiply the two denominators. $z = \dfrac{ -7(x + 6) \times (x + 9)(x - 1) } { (x + 9) \times 3(x + 6) } $ $z = \dfrac{ -7(x + 6)(x + 9)(x - 1)}{ 3(x + 9)(x + 6)} $ Notice that $(x + 6)$ and $(x + 9)$ appear in both the numerator and denominator so we can cancel them. $z = \dfrac{ -7\cancel{(x + 6)}(x + 9)(x - 1)}{ 3\cancel{(x + 9)}(x + 6)} $ We are dividing by $x + 9$ , so $x + 9 \neq 0$ Therefore, $x \neq -9$ $z = \dfrac{ -7\cancel{(x + 6)}\cancel{(x + 9)}(x - 1)}{ 3\cancel{(x + 9)}\cancel{(x + 6)}} $ We are dividing by $x + 6$ , so $x + 6 \neq 0$ Therefore, $x \neq -6$ $z = \dfrac{-7(x - 1)}{3} ; \space x \neq -9 ; \space x \neq -6 $